I'm a little confused about situations where multiple chemical compounds can be reduced/oxidized and the likelihood of this happening to one compound over the others, depending on the electrode potential of that compound. Here are some exercises I'm trying to figure out on this topic and their answers as given to me:

First exercise: electrolysis with inert electrodes and aqueous solution $\ce{CuI_2}$, the question is: "why do we observe $\ce{I_2}$ gas release rather than $\ce{O_2}$".

The answer is: "If $\ce{H_2O}$ is oxidized at the anode (to form $\ce{O_2}$) rather than $\ce{I^-}$ (to form $\ce{I_2}$), we have $E^°_{cell}$ = $E^°_{Cu^{++}/Cu}$ - $E^°_{H_2O/O_2} = 0.337 - 1.23 = -0.893V$ which is lower than $E^°_{cell}$ = $E^°_{Cu^{++}/Cu}$ - $E^°_{I_2/I^-} = 0.337 - 0.5355 = -0.1985V$ and thus, since the potential is higher (in absolute value) for water, no $\ce{O_2}$ should be released as long as their is $\ce{I^-}$ ions available"

Second exercise: electrolysis with inert electrodes and aqueous solution at $\ce{pH = 0}$, $\ce{T=298,15K}$, $\ce{p=1bar}$ and containing $10^{-2}M$ of $\ce{Ag^+}$ ions and $10^{-2}M$ of $\ce{Cu^{++}}$ ions, the question is: "discuss the possibility of having a copper deposit without impurities from the silver".

The answer is: "[I'm skipping some calculations but it can be provided if required] We get the standard electrode potentials: $E^°_{Cu^{++}/Cu} = 0.337V$ and $E^°_{Ag^+/Ag}$ = $0.799V$ and from Nernst equation (so calculation of the actual -non standard- cell potential) we get for $\ce{H_2O/Ag}$ and $\ce{H_2O/Cu}$, respectively: $E_{cell}$ = $- 0.549V$ and $E_{cell}$ = $- 0.952V$ then we calculate the concentration of $\ce{Ag^+}$ at $- 0.952V$ (using Nernst law again) and we get: $[\ce{Ag^+}] = 1,7\times10^{-9}M$ , which is very low compared to [$\ce{Cu^{2+}}$] = $10^{-2}M$, so almost only $\ce{Cu}$ will deposit on the electrode."

So here is my question on all this: I have read everywhere that "more negative electrode potential means more likely to be oxidised", which actually correspond correctly to my first exercise ($E^°_{I_2/I^-} = 0.5355V$ is "more negative" than $E^°_{H_2O/O_2} = 1.23V$), is it correct to apply this reasoning here though? And also that "less negative electrode potential means more likely to be reduced" and there I have a problem, in the second exercise $E^°_{Cu^{++}/Cu} = 0.337V$ is not "less negative" than $E^°_{Ag^+/Ag} = 0.799V$ but quite the contrary, still we calculated that almost no $\ce{Ag^+}$ was deposited (so reduced, to my understanding) compared to $\ce{Cu^{++}}$.

Could you help me to understand why we get this ? Simple and general guidelines for this kind of reasonings would really help.

EDIT (for those who would try to understand this topic): As Robert DiGiovanni pointed out (see the discussion below his message), the conclusion for the second exercise is incorrect. Since $E^°_{Ag^+/Ag}$ is more positive than $E^°_{Cu^{++}/Cu}$, $\ce{Ag}$ will reduce first. Furthermore, the result from the Nernst equation actually confirms the fact that copper cannot be deposited pure. Since $\ce{[Ag^+]}$ is found really low from Nernst equation (giving the cell potential at equilibrium), it means that most $\ce{Ag^+}$ will be reduced when reaching equilibrium, and thus most of the Ag will be deposited on the electrode.

  • $\begingroup$ Convenient reference for text/formula formatting: Notation basics / Formatting of math/chem expressions / upright vs italic // For more: Math SE MathJax tutorial. // Not to be applied in CH SE titles. $\endgroup$
    – Poutnik
    Commented Dec 28, 2022 at 6:23
  • $\begingroup$ $\ce{2 Cu^2+(aq) + 4 I-(aq) -> 2 CuI(s) + I2(aq,s)}$ $\endgroup$
    – Poutnik
    Commented Dec 28, 2022 at 6:29
  • $\begingroup$ You should ask the author of the task to show you $\ce{CuI2}$ solution. $\endgroup$
    – Poutnik
    Commented Dec 28, 2022 at 7:16
  • $\begingroup$ @Poutnik what copper gets reduced to is irrelevant. Hydrolysis of water, written as suggested is H20 ---> H2 + O2 - 1.23 V. Before this happens, 2 I- ----> I2 + 2e- - 0.54 V. Looking further, we find Cu++ + 2 e- ---> Cu + 0.34 V and Cu+ + 1 e- ---> Cu + 0.52 V. It seems metallic Cu would be formed at the cathode. $\endgroup$ Commented Dec 28, 2022 at 9:31
  • $\begingroup$ @RobertDiGiovanni It is relevant what iodide is oxidized to. It does does get chance to get electrooxidized in priority when it is already oxidized. Similarly for copper. CuI precipitate is not easy to reduce on cathode. $\endgroup$
    – Poutnik
    Commented Dec 28, 2022 at 9:49

2 Answers 2


The main theme of the questions is a very good academic exercise but the question writer (whoever wrote the the two questions for students) made them vague and the answers are more convoluted than the question. It is bound to confuse students.

For the first question, $\ce{CuI2}$ does not exist and it cannot exist in solution. Copper(II) is a strong oxidizing agent that will oxidize iodide to free iodine. Let us correct the question and say, if we had a solution of $\ce{KI}$ i.e., potassium iodide in water and if we electrolyze it, what products do we expect at the anode? The general theme for addressing such questions would be:

  1. List all the possible ions and molecules in the solution: We have $\ce{H2O}$, free $\ce{K+}$ and $\ce{I-}$, $\ce{H+}$ and $\ce{OH-}$

  2. Since we are interested in oxidation only, we will think about the species which can be oxidized at the anode, the only chemically realistic possibilities are that $\ce{H2O}$ and $\ce{I-}$. Forget the cathode for the time being.

  3. Now check, which oxidation is thermodynamically favored by looking up the electrode potential tables. All modern electrode potential tables are written as reduction potentials. Note that all listed electrode potentials are provided under standard conditions, if they have $\ce{H+}$ or $\ce{OH-}$ in the equations, then under standard conditions, we mean that they have unit activity or say pH=0 or pH=14 respectively.

Please do not flip signs of electrode potentials. They are sign invariant. This action is frowned upon by modern electrochemists, although some general chemistry textbooks still do that.

$$\ce{I2 + 2e- ⇌ 2 I- (+0.5355 V)}$$


$$\ce{O2(g) + 4H+(aq) + 4e- ⇌ 2H2O(l) (+1.229 V)}$$

If we interpret the above equations as follows given that we have $\ce{H2O}$ and $\ce{I-}$, then at the anode, it will require less energy to oxidize iodide ion to free iodine vs. water to free oxygen. Thus at the anode, thermodynamic oxidation of iodide will be favored instead of water.

As to the second question, the logic in the answer there is also convoluted. It will be far easier to consider cathode reactions only. You have to consider reductions only this time. Follow the steps above list the reduction reactions under non-standard conditions. You may post the second question separately after attempting along similar lines of reasonings.

  • $\begingroup$ I personally found that flipping sign to make it match the actual direction of the reaction was really convenient, because you just have to find "which potential is more positive" to find which reaction is favored. Otherwise, you need different rules (more/less ... means more likely to reduce/oxidise), I find it more confusing and harder to see what's actually going on. Also, flipping sign if needed tells you some logical information: if sign is negative the reaction requires the application of a potential to occur, if sign is positive the reaction is spontaneous and produces a potential. $\endgroup$
    – c.leblanc
    Commented Dec 28, 2022 at 17:15
  • $\begingroup$ I agree, this is how it is taught, but I have personally discussed this with three world's leading electrochemists and they do not agree with this sign flipping. Convenience does not mean scientific correctness. Hope the general guidelines helped. $\endgroup$
    – ACR
    Commented Dec 28, 2022 at 17:37
  • $\begingroup$ @Achem I frequently compare it to analogy "Left river bank is on your left(naive approach)" versus "left river bank is on your left when looking downstream(geography)". And the latter ported to electrochemistry - it is related to Gibbs energy of reaction oxidating H2 to H+ in SHE (and not reversed). $\endgroup$
    – Poutnik
    Commented Dec 28, 2022 at 19:37
  • $\begingroup$ @Poutnik, Good analogy but those electrrochemists have a different reasoning. Note that it is okay to reverse signs in thermodynamics but not in electrostatics. These electrochemists told me that it is an electrostatic sign associated with the electrode potential when written in reduction form- this is the true electrostatic sign of the electrode with reference to SHE, hence invariant. I cannot call a proton as a negatively charged particle, no matter how we look at it. $\endgroup$
    – ACR
    Commented Dec 28, 2022 at 20:27
  • $\begingroup$ @AChem In a way, linking electrochemical potential to flipping Delta G of reversible reactions was not the best move. $\endgroup$
    – Poutnik
    Commented Dec 28, 2022 at 20:33

Iodide gives up its electron easier than water, silver accepts electrons before copper.

The best way is to draw the half reactions in the direction they occur. Then the math is easy. For example:

$$\ce{2Na metal + Cl2 -> 2 NaCl}$$

The oxidation half reaction is: $$\ce{2Na -> 2Na+ + 2e- + 2.71 V}$$

You may find it on a Standard Reduction Potential Chart as:

$$\ce{ Na+ + e- -> Na - 2.71 V}$$

but that is not the direction the reaction is running.

Similarly, the reduction half reaction is:

$$\ce{Cl2 + 2e- -> 2 Cl- + 1.36 V}$$

2.71 + 1.36 = + 4.07 V. Half reaction redox potentials are simply added.

Just remember to set your redox half reactions up this way and the comparisons will be easy.

Background reading on Potentiometric Titrations may be helpful.

  • $\begingroup$ So, if you want to plate out pure copper, plate out silver first, then replace the cathode. $\endgroup$ Commented Dec 28, 2022 at 10:13
  • 1
    $\begingroup$ Reactions that are not net positive can be run by applying potential from an outside source. The rule of thumb is higher positive values for drawn half reactions happen first under all conditions. This just in: Cu ++ ---> Cu + 0.34 V. Cu ++ ---> Cu + + 0.15 V. $\endgroup$ Commented Dec 28, 2022 at 10:48
  • 1
    $\begingroup$ @c.leblanc. draw the half reactions in the order they proceed. The more positive ones go first. Discuss this with your professor and others. Incorrect test answers can be very confusing. $\endgroup$ Commented Dec 28, 2022 at 11:19
  • 1
    $\begingroup$ @c.leblanc. from the Nernst equation, what we get is at that voltage most of the silver is already reduced. Think titration. If you apply voltage to a copper and silver ion solution, the silver will ground the current is it reduces, showing a "plateau" until the silver ions are reduced. Then voltage increases until copper grounds at the higher potential. I had a great professor. Very lucky! $\endgroup$ Commented Dec 28, 2022 at 11:33
  • 1
    $\begingroup$ I was actually thinking the same. Since we get that it is the silver ions that are really low in concentration, I guess it does make sense that it actually means they reduced almost fully (so silver deposit would be high) at equilibrium (since Nernst equation represent the potential of the reaction at equilibrium). $\endgroup$
    – c.leblanc
    Commented Dec 28, 2022 at 12:11

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